Question 901911
<pre>
2cos²(3x) + 5cos(3x) - 3 < 0

The left factors just as 2u²+5u-3 = (u+3)(2u-1)

[cos(3x)+3][2cos(3x)-1] < 0

We find the critical values by finding the zeros of the left side

The first factor has no zeros since no cosine can be -3
The second factor has zeros when cos(3x) = {{{1/2}}}

This is when {{{3x=pi/3 +2n*pi}}}   and when {{{3x=5pi/3+2n*pi}}}
             {{{x=pi/9 +2n*pi/3}}}          {{{x=5pi/9+2n*pi/3}}} 
             {{{x=pi/9 +6n*pi/9}}}          {{{x=5pi/9+6n*pi/9}}}
             {{{x=expr(pi/9)(1+6n)}}}            {{{x=expr((5pi)/9)(5+6n)}}}

{{{graph(400,400,-2pi,2pi,-7,5,2cos(3x)^2+5cos(3x)-3)}}}

From the graph of y = 2cos²(3x) + 5cos(3x) - 3 we see that the graph is negative
(below the x-axis) in all open intervals
{{{(matrix(1,3,expr(pi/9)(1+6n),",",expr((5pi)/9)(5+6n)))}}}, when n is any integer.       

Edwin</pre>